Respuesta :
Explanation:
It is known that for first order reaction, the equation is as follows.
t = [tex]\frac{2.303}{K} log \frac{[C_{4}H_{8}]_{o}}{[C_{4}H_{8}]_{t}}[/tex]
t = ?, K = rate constant = 79 1/s
Initial conc. of [tex]C_{4}H_{8}[/tex] = 1.68
Decompose amount of [tex]C_{4}H_{8}[/tex] = 52% of 1.68
= [tex]\frac{52}{100} \times 1.68[/tex]
= 0.8736
= 0.87
Now, [tex][C_{4}H_{8}]_{t}[/tex] = (1.68 - 0.87)
= 0.81
Therefore, calculate the value of t as follows.
t = [tex]\frac{2.303}{K} log \frac{[C_{4}H_{8}]_{o}}{[C_{4}H_{8}]_{t}}[/tex]
= [tex]\frac{2.303}{79 s^{-1}} log \frac{1.68}{0.81}[/tex]
= [tex]\frac{2.303}{79 s^{-1}} \times 0.316[/tex]
= [tex]9.212 \times 10^{-3}[/tex] s
= [tex]0.00921 s[/tex]
Thus, we can conclude that 0.00921 s will be taken for 52% of the cyclobutane to decompose.
